Question

In the one-dimensional motion of a particle, the relation between position x and time t is given by
x2 + 2x = t (here x > 0). Choose the correct statement :
(A) The retardation of the particle is 1/4(x+1)^3
(B) The uniform acceleration of the particle is
1/(x + 1)^3
(C) The uniform velocity of the particle is 1/(x+1)^3
(D) The particle has a variable acceleration of 4t + 6.​

Answer 1

Answer:

Option (a) the retardation is 1/4(x+1)^3

Explanation:

The give relation between position x and time t is

$x^2+2x=t$

We know that rate of change of displacement is equal to velocity

i.e. dx/dt = v

And, rate of change of velocity is acceleration

i.e. dv/dt = a

which can be written as

$a=\frac{dv}{dt}$

or, $a=\frac{dv/dx}{dt/dx}$

or, $a=v\frac{dv}{dx}$

Differentiating the given relation in x and t, with respect to t

$2x\frac{dx}{dt} +2\frac{dx}{dt}=1$

or, $\frac{dx}{dt}=\frac{1}{2(x+1)}$

or, $v=\frac{1}{2(x+1)}$

Again differentiating the above w.r.t. x

$\frac{dv}{dx}=-\frac{1}{2(x+1)^2}$

but

$a=v\frac{dv}{dx}$

$\implies a=v\frac{dv}{dx}=-\frac{1}{2(x+1)^2}\times \frac{1}{2(x+1)}$

$\implies a=-\frac{1}{4(x+1)^3}$

The negative sign indicates retardation

Therefore, the retardation is 1/4(x+1)³